A frictionless inclined plane of length l having inclination `theta` is placed inside a lift which is accelerating downward with on acceleration a `(ltg)`. If a block is allowed to move down the inclined plane, from rest, then the time taken by the block to slide from top of the inclined plane to the bottom of the inclined plane is

Explanation

To solve the problem of a block sliding down a frictionless inclined plane inside a downward-accelerating lift, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The block experiences gravitational force \( mg \) acting downward.
- The lift is accelerating downward with acceleration \( a \) (where \( a < g \)).
- The effective gravitational force acting on the block in the lift's frame is \( mg - ma = m(g - a) \).

2. **Resolve the Effective Gravitational Force:**
- The effective gravitational force can be resolved into two components relative to the inclined plane:
- **Perpendicular to the incline:** \( (g - a) \cos \theta \)
- **Parallel to the incline:** \( (g - a) \sin \theta \)

3. **Determine the Acceleration of the Block:**
- The acceleration \( a_{\text{incline}} \) of the block down the incline is given by the component of the effective gravitational force acting parallel to the incline:
\[ a_{\text{incline}} = (g - a) \sin \theta \]

4. **Use the Kinematic Equation to Find Time:**
- The block starts from rest and slides down the incline of length \( L \). We can use the kinematic equation:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s = L \) (the length of the incline),
- \( u = 0 \) (initial velocity),
- \( a = a_{\text{incline}} = (g - a) \sin \theta \).

Substituting these values into the equation gives:
\[ L = 0 \cdot t + \frac{1}{2} (g - a) \sin \theta \cdot t^2 \]
Simplifying, we have:
\[ L = \frac{1}{2} (g - a) \sin \theta \cdot t^2 \]

5. **Solve for Time \( t \):**
- Rearranging the equation to solve for \( t^2 \):
\[ t^2 = \frac{2L}{(g - a) \sin \theta} \]
- Taking the square root to find \( t \):
\[ t = \sqrt{\frac{2L}{(g - a) \sin \theta}} \]

### Final Answer:
The time taken by the block to slide from the top of the inclined plane to the bottom is:
\[ t = \sqrt{\frac{2L}{(g - a) \sin \theta}} \]